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Prove/Disprove:

Let $R$ be a commutative ring with unity. In $R$ for any two distinct non-trivial ideals $I,J\subseteq R$ we have $(I+J)^2=I+J$. Given ideals $I$ and $J$ in $R$ can we find idempotents $e,f\in R$ such that $I=Re,J=Rf$ where $ef=0$?

An ideal $I$ is said to be non-trivial if $I\neq \{0\},R.$

MY TRY:

If I try to prove the fact then let us assume that $I,J$ be two non-trivial ideals in $R$,then $(I+J)^2=I+J$,How should I show that $I=Re,J=Rf$ where $e^2=e;f^2=f;ef=0$??

Learnmore
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    The question is ill-posed. You introduce a ring with a property and then you ask something about symbols $I$ and $J$ that are not defined any more. Also, in the property you describe for $R$, you do not exclude the possibility that $I=J$. In this case, even if $I$ is generated by an idempotent, $I=J=Re$ and $e^2=e\ne 0$ because $I$ was assumed nonzero. I feel like you might have misphrased something. – Jesko Hüttenhain Jun 04 '17 at 08:12
  • I don't understand why the question is ill-posed? Why is $I,J$ not defined?@JeskoHüttenhain – Learnmore Jun 04 '17 at 10:25
  • Yes,check the edits @Arthur – Learnmore Jun 04 '17 at 10:26
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    The question is ill posed because logically, there is no connection between the $I$ and $J$ of the two paragraphs. The way the first paragraph is written implies that the letters $I$ and $J$ are dummies used to describe a property of $R$, and that you're done with them the moment that definition is finished. That is why $I$ and $J$ are undefined in the second paragraph. – Arthur Jun 04 '17 at 12:20
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    No, you've just made the two paragraphs into one. That doesn't change the logic of what's going on. You might as well write "Let $R$ be a commutative ring with unity. For any two distinct non-trivial ideals $I,J\subseteq R$ we have $(I+J)^2=I+J$. Does there exist idempotents $e,f\in R$ such that $K=Re,L=Rf$ where $ef=0$?" You have to tell us what $I$ and $J$ are in the second sentence as well, and making that period into a comma is not enough. – Arthur Jun 04 '17 at 17:45
  • Question Edited;$I,J$ are two given ideals in the ring $R$ in which the given property holds@Arthur – Learnmore Jun 05 '17 at 04:19

3 Answers3

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I suggest you the following:

1) Assume that $R$ is a domain. What are the idempotents of $R$ ?

2) Pick your favorite domain $R$ and nontrivial ideals $I,J$ such that $I+J=R.$

3) Conclude.

GreginGre
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  • Hard to understand your answer!Not sure what you wanted to say at all – Learnmore Jun 04 '17 at 14:15
  • @RishabhPant The answer points to the fact that the answer to your question is no. Example: $R=\mathbb{Z}$, $I=2\mathbb{Z}$ and $J=3\mathbb{Z}$. – egreg Jun 04 '17 at 14:58
  • I said for any two ideals not just a particular one @egreg; take $I=2\Bbb Z,J=4\Bbb Z$ – Learnmore Jun 04 '17 at 16:24
  • @RishabhPant Then $I+J=2\mathbb{Z}\ne(I+J)^2=4\mathbb{Z}$. Anyway, a single counterexample disproves a general statement like the one you have in your question, doesn't it? – egreg Jun 04 '17 at 16:31
  • I am still unable to understand your point @egreg; I said that for any two distinct ideals $I,J$ we have $(I+J)^2=I+J$ but the hypothesis fails if you take $R=\Bbb Z$ – Learnmore Jun 05 '17 at 04:14
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OK, I misunderstood you question. You want $(I+J)^2=I+J$ for every pair of non trivial distinct ideals.

The answer is still NO. Take $R=\mathbb{C}^3$. Any ideal of $R$ is the direct product of copies of $(0)$ or $\mathbb{C}$, so $R$ clearly satisfies your assumption.

Take $I=\mathbb{C}\times \mathbb{C}\times (0)$ and $J=\mathbb{C}\times (0)\times (0).$

Then $I=(1,1,0)R$ and $J=(1,0,0)R$. Notice that the idempotents of $R$ are the vectors whose coordinates are $0$ or $1$. There is only one idempotent generating $I$ , which is $e=(1,1,0)$ and there is only one idempotent generating $J$, which is $f=(1,0,0)$. Unfortunately $ef\neq 0$.

GreginGre
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The current version of the question is:

Let $R$ be a commutative ring with unity. In $R$ for any two distinct non-trivial ideals $I,J\subseteq R$ we have $(I+J)^2=I+J$. Given ideals $I$ and $J$ in $R$ can we find idempotents $e,f\in R$ such that $I=Re,J=Rf$ where $ef=0$?

An ideal $I$ is said to be non-trivial if $I\neq \{0\},R.$

The answer is no:

Let $R$ be a field.

The condition "for any two distinct non-trivial ideals $I,J\subseteq R$ we have $(I+J)^2=I+J$" is vacuously satisfied be cause there are no non-trivial ideals.

The condition "given ideals $I$ and $J$ in $R$ we can find idempotents $e,f\in R$ such that $I=Re,J=Rf$ where $ef=0$" does not hold because we can set $I=J=(1)$.

Pierre-Yves Gaillard
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